LMX2595的datasheet中给出的自动斜坡例子中,RAMP0_INC =13422, 根究公式可得
| (2000 MHz) / (50 MHz) × 2^24/ 50000 = 13422 |
此公式的具体解释是什么?
具体斜坡的计算:斜坡时长和斜坡△f怎样设计计算的?
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LMX2595的datasheet中给出的自动斜坡例子中,RAMP0_INC =13422, 根究公式可得
| (2000 MHz) / (50 MHz) × 2^24/ 50000 = 13422 |
此公式的具体解释是什么?
具体斜坡的计算:斜坡时长和斜坡△f怎样设计计算的?
您好,根据数据手册中的例子,Suppose user wants to generate a sawtooth ramp that goes from 8 to 10 GHz in 2 ms (including calibration breaks) with a phase-detector frequency of 50 MHz. Divide this into segments of 50 MHz where the VCO ramps for 25 µs, then calibrates for 25 µs, for a total of 50 µs. There would therefore be 40 such segments which span over a 2-GHz range and would take 2 ms, including calibration time.
首先扫频范围为2GHz=2000Mhz,2ms内完成扫频。fpd=50Mhz,如果每50us扫频一次。
这样理解的话,2ms完成,一共需要2ms/50us次的扫频。 那么2000Mhz的扫频范围话,
ramp increment即RAMPx_INC=2000Mhz/2ms/50us.
我是这么理解的,和数据手册的公式有出入,我再想一下。
您 的问题我已经在下面链接中给出回答,后续有什么问题,我会在这个链接中跟进。